Mathematical inquiry processes: Notice patterns and connections; conjecture, generalise and prove. Conceptual field of inquiry: Laws of surds; multiplication and simplification of surds; Pythagoras' theorem; and geometric sequences.
Zack Miodownik, a teacher of mathematics in London (UK), created the prompt for his year 9 class. The students had studied Pythagoras' theorem and Zack used the prompt to introduce the class to surds.
Students' responses from the question, notice, wonder phase include:
The numbers on the short sides of the triangles are reversed.
One number is three times the other one.
As they are right-angled triangles, you can use Pythagoras' theorem: The length of the hypotenuse equals the square root of the sum of the squares of the two short sides.
How do you square a term with a surd?
The square root of any number squared is the number, for example √(6) x √(6) = 6.
Why are some triangles missing, such as the one for 3√(9) and 9√(3)?
Which is the longest short side, the base or the height?
Do the sides form a sequence from triangle to triangle? Is it a geometric sequence?
As students feed back, the teacher can formalise the mathematical language by defining a surd.
At this point, the teacher should point out the non-examples that arise if the triangles with base lengths 3√(9) and 4√(12) are included in the prompt. As 3√(9) = 9 and 12√(4) = 24, neither is an irrational number and, therefore, not a surd. We can speculate that the sides of the triangles in the prompt have to be surds.
During a discussion of the longest short side in each right-angled triangle, the teacher can start to introduce the laws of surds, such as √(ab) = √(a) x √(b), at a relevant and meaningful point in the inquiry.
In the first triangle the longer of the two short sides is 6√(2) because 6√(2) = 2 x 3 x √(2), while 2√(6) = 2 x √(3) x √(2), and 3 > √(3).
The same law arises when explaining why the triangle with base length 6√(18) is also missing from the prompt. As 6√(18) can be simplified to 18√(2), the length of a short side would not be in its simplest form.
Thus, we conclude that permissible triangles in the context of the inquiry are those with surds that cannot be simplified. The next case after those in the prompt has base length 10√(30).
As students square the lengths of the short sides and square root their sum, they begin to identify a pattern: (2√(6))2 + (6√(2))2 = √(24+72) = √(96) = 4√(6). In the first case the hypotenuse is twice the length of the shortest short side. Is that true for all cases?
Zack designed a four-part sheet for his students to structure their inquiry (included in the slides). Alternatively, experienced inquirers might use the regulatory cards to decide how to proceed with the inquiry.
November 2025
In a structured inquiry students explore more cases in which the lengths of the two short sides, in general terms, are n√(3n) and 3n√(n). This includes the three cases in the prompt and others that are 'missing' in between them.
Students form the generalisation that the length of the hypotenuse is twice as long as the length of the shortest short side. The teacher might then co-construct a proof (see the illustration below) as a model for students to follow in subsequent lines of inquiry.
The first line of inquiry takes students through the case of k = 3. What if k = 2, 4, or 5? For example, when k = 2, the length of the short sides would be in the form n√(2n) and 2n√(n). Is the length of the hypotenuse still twice as long as the length of the shortest short side?
Students might form a generalisation for each value of k and attempt to prove for the individual cases before going on to prove a generalisation for all values of k (see the illustration below).
Students can change the prompt and generate their own example sets.
What if both short sides of the right-angled triangles are the same? If the lengths are 6√(2), for example, is there a connection with the length of the hypotenuse?
What if we change the lengths of the short sides to 6 - √(2) and 6 + √(2)? Is it possible to generalise for a - √(b) and a + √(b) and prove?
See the mathematical notes.
In response to the final one of the students' questions about the prompt - do the lengths of the bases form a geometric sequence? - Zack designed a task as part of another line of inquiry. The task requires students to find the common ratio before filling in the gaps to complete a sequence. They can then make up their own sequences.
In introducing the task, the teacher draws on students' existing knowledge to explain why the lengths of the bases do not form a geometric sequence. As 5√(15) ÷ 2√(6) ≠ 7√(21) ÷ 5√(15), there is not a common ratio between the terms.